The pressure of sea water increases by 1.0atm for each 10m increase in the depth, by what percentage is the density of water increased in the deepest ocean of water of 12km. Compressibility is 5.0×10^-5 atm

The percentage by which the water density increased is 4.1 %

The known values are;

The increase in pressure per 10 meter increase in depth = 1.0 atm

The depth of the deepest ocean = 12 km = 12,000 m

The compressibility of the ocean = 5.0 × 10⁻⁵ 1/atm

The unknown

The percentage the density of water increased in the deepest ocean

Strategy;

Find the pressure at the deepest point of the deepest ocean and apply the compressibility

We have;

The change in pressure, = (12,000 m/(10 m)) × 1.0 atm = 1,200 atm

Therefore, we have for one cubic meter of water

Therefore;

V = 5.0 × 10⁻⁵ atm⁻¹ × 1 m³ × 1,200 atm = 0.06 m³

The new volume = V - V

∴ The new volume = 1 m³ - 0.06 m³ = 0.94 m³

The initial density = mass/(1 m³)

The new density = mass/(0.96 m³)

The percentage increase in density, ρ%, is given as follows;

∴  ρ% =  4.1 %

The percentage by which the water density increased, ρ% = 4.1 %

Learn more about compressibility here;

the inner core.

good luck: )

klicker